Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Provide solution for RD Sharma maths class 12 chapter Differential Equations exercise 21.11 question 10

Provide solution for RD Sharma maths class 12 chapter Differential Equations exercise 21.11 question 10

Answers (1)

Answer: i=\frac{E}{R}\left[1-e^{\left(-\frac{K}{L}\right) t}\right]

Given:  L \frac{d i}{d t}+R i=E

To find: To prove that i=\frac{E}{R}\left[1-e^{\left(-\frac{R}{L}\right) t}\right]

Hint: Integrate the linear differential equation to find integral factor.

Solution: L \frac{d i}{d t}+R i=E

 Dividing by L

        =\frac{d t}{d t}+\frac{R}{L} i=\frac{E}{L}

This is a linear differential equation and comparing it with

        =\frac{d y}{d x}+P y=Q

Such that,

        =P=\frac{R}{L^{\prime}}, Q=\frac{E}{L}

We know that

        \begin{aligned} &\text { I.F. }=e^{\int p d t} \\\\ &=e^{\int \frac{R}{L} d t} \\\\ &=e^{\left(\frac{R}{L}\right) t} \end{aligned}

So, the solution is given by

        \begin{aligned} &i(I f)=\int Q(I f) d t+C \\\\ &=>i\left[e^{\left(\frac{R}{L}\right) t}\right]=\int \frac{E}{L}\left[e^{\left(\frac{R}{L}\right) t}\right] d t+C \\\\ &=>i\left[e^{\left(\frac{R}{L}\right) t}\right]=\frac{E}{L} \times \frac{L}{R}\left[e^{\left(\frac{R}{L}\right) t}\right]+C \end{aligned}

        \begin{aligned} &=>i\left[e^{\left(\frac{R}{L}\right) t}\right]=\frac{E}{R}\left[e^{\left(\frac{R}{L}\right) t}\right]+C \\\\ &=>i=\frac{\frac{E}{R}\left[e^{\left(\frac{R}{L}\right) t}\right]+C}{\left[e^{\left.\left(\frac{R}{L}\right) t\right]}\right.} \end{aligned}

        \begin{aligned} &\text { => } i=\frac{E}{R}+\frac{C}{\left[e^{\left(\frac{R}{L}\right) t}\right]} \\\\ &=>i=\frac{E}{R}+C\left[e^{-\left(\frac{R}{L}\right) t}\right] \ldots(i) \end{aligned}

Initially no current passes through the circuit so,i=0, t=0

        \begin{aligned} &=>0=\frac{E}{R}+C e^{0} \\\\ &=>0=\frac{E}{R}+C \\\\ &=>C=-\frac{E}{R} \end{aligned}

Substituting values of C in equation (i)

        \begin{aligned} &=>i=\frac{E}{R}-\frac{E}{R}\left[e^{-\left(\frac{R}{L}\right) t}\right] \\\\ &=>i=\frac{E}{R}\left[1-e^{-\left(\frac{R}{L}\right) t}\right] \end{aligned}

 

Posted by

infoexpert26

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Trending Articles/News

anam-khan

Board Exam Time Table 2025 Live: UP Board exams from February 24; CBSE, BSEB date sheet soon
anam-khan

CBSE Date Sheet 2025 Live: Class 10, 12 board exam dates soon; syllabus, sample papers
anam-khan

Extra time in CBSE exams for students with type 1 diabetes: Kerala SHRC seeks report on plea

Latest Question